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Đề đánh bị lỗi.
Áp dụng bất đẳng thức Bunhiacopski:
\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{\left[\sqrt{c}^2+\sqrt{\left(a-c\right)}^2\right]\left[\sqrt{c}^2+\sqrt{\left(b-c\right)}^2\right]}\)
\(=\sqrt{\left(c+a-c\right)\left(c+b-c\right)}=\sqrt{ab}\)
Vì a ; b ; c dương , áp dụng BĐT Cô - si cho các cặp số dương , ta có :
\(\frac{c}{b}+\frac{a-c}{a}\ge2\sqrt{\frac{c\left(a-c\right)}{ab}}\)
\(\frac{c}{a}+\frac{b-c}{b}\ge2\sqrt{\frac{c\left(b-c\right)}{ab}}\)
\(\Rightarrow2\ge2\sqrt{\frac{c\left(a-c\right)}{ab}}+2\sqrt{\frac{c\left(b-c\right)}{ab}}\)
\(\Rightarrow1\ge\frac{\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}}{\sqrt{ab}}\)
\(\Rightarrow\sqrt{ab}\ge\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{c}{b}=\frac{a-c}{a};\frac{c}{a}=\frac{b-c}{b}\)
\(\Leftrightarrow\frac{c}{b}+\frac{c}{a}=1\) \(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\)
Vì \(a;b\ge c\Rightarrow a=b=2c\)
Vậy ...
BĐT cần chứng minh tương đương: \(\sqrt{\frac{c\left(a-c\right)}{ba}}+\sqrt{\frac{c\left(b-c\right)}{ab}}\le1\)
Áp dụng BĐT Cauchy:
\(VT\le\frac{1}{2}\left(\frac{c}{b}+\frac{a-c}{a}+\frac{c}{a}+\frac{b-c}{b}\right)=\frac{1}{2}\left(\frac{a-c+c}{a}+\frac{c+b-c}{b}\right)=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=2c\)
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sửa đề\(\frac{1}{x^2+1}+\frac{1}{y^2+1}\ge\frac{2}{1+xy}\)
\(\Leftrightarrow\frac{1}{x^2+1}+\frac{1}{y^2+1}-\frac{2}{1+xy}\ge0\)
\(\Leftrightarrow\left(\frac{1}{1+x^2}-\frac{1}{1+xy}\right)+\left(\frac{1}{1+y^2}-\frac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\frac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\frac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\frac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)( luôn đúng với \(x,y\ge1\))
Đpcm
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
tìm trc khi hỏi Câu hỏi của Hoàng Thiên - Toán lớp 9 - Học toán với OnlineMath
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